41st VTRMC, 2019, Solutions

  1. Let M denote the minimal value of f (n). Clearly M≤2 + 7 + 7 + 1 = 17. We will show that M = 17, so assume by way of contradiction that M < 17. Choose n∈ℕ with f (n) = M, and write n in reverse order as 1a1...ad where ad≠ 0 (so n is a (d + 1)-digit number). We have f (n)≡2771n≡(-1)nmod 9. First assume that n is odd, so f (n)≡ -1 mod 9, so we must have a1 + ... + ad = 7. We also have 1 -a1 + a2 - ...≡2771n≡ -1 mod 11, so -a1 + a2 -a3 + ... = -2. Adding these two equations, we obtain 2a2 +2a4 + ... = 5, a contradiction because the left hand side is an even integer and the right hand side is an odd integer. Now assume that n is an even integer. Then we have f (n)≡1 mod 9 and therefore a1 + ... + ad = 9. Also 1 -a1 + a2 - ...≡1 mod 11 and therefore -a1 + a2 -a3 + ... = 0. Adding the last two equations, we obtain 2a2 +2a4 + ... = 9, again a contradiction and the result follows.

  2. Since BX/XA = 9, we see that AX = AB/10 and we deduce that the area of AXC is 1/10 of the area of ABC, because they have the same height. Using the fact that the area of XYC is 9/100 of the area of ABC, we find that the area XYB is 81/100 of the area of ABC. Therefore the area of XBY is 9/10 of the area of XBC. Let H be the point on AB such that AHC = 90o. Since XBY and XBC have the same base, we see that MY = (9/10)CH. Now MBY and HBC are similar, consequently

    HB = (10/9)MB = (10/9)·(1/2)·XB = (10/9)·(1/2)·(9/10)AB = (1/2)AB.

    Therefore AC = BC and hence BC = 20.

  3. Define g(x) = ∫0x(1 -t)f (tdt. Then g(0) = 0 and

    g(1) = ∫01f (xdx - ∫01xf (xdx    
      = ∑d=0nad/(d + 1) - ∑d=0nad/(d + 2)    
      = ∑d=0nad/((d + 2)(d + 1)) = 0.    

    By Rolle's theorem, there exists q∈(0, 1) such that g'(q) = 0, that is (1 -q)f (q) = 0. Since q≠1, we deduce that f (q) = 0 as required.

  4. Let I = ∫01x2/(x + √(1 -x2)) dx. We make the substitution x = sin t. Then dx = dt cos t and we see that I = ∫0π/2(sin2t cos t)/(sin t + cos tdt. Also by making the substitution x = cos t, we see that I = ∫0π/2(cos2t sin t)/(sin t + cos tdt and we deduce that

    2I = ∫0π/2(sin2t cos t + cos2t sin t)/(sin t + cos tdt.

    Since sin2t cos t + cos2t sin t = sin t cos t(sin t + cos t), we find that 2I = ∫0π/2sin t cos t dt. Therefore 4I = ∫0π/2sin 2t dt and we conclude that I = 1/4.

  5. We make the substitution t = 1/x. Let y' and y'' denote the first and second derivatives of y with respect to t, respectively.Then dy/dx = -t2y and d2y/dx2 = 2t3y' + t4y'' and by substituting back into the original equation, we obtain y'' + (2t-1 -2)y' + (1 - 2t-1)y = 0. It is easy to see that y = et is a solution to this equation. We now use reduction of order to obtain a second solution, so let y = f (t)et be another solution, where f is to be determined. Then f'' + 2t-1f' = 0, which has the solution f = t-1. We deduce that e1/x and xe1/x are solutions to the original equation. Since these solutions are clearly linearly independent, it follows that the general solution to the original equation is y = C1e1/x + C2xe1/x, where C1 and C2 are arbitrary constants.

  6. For each sS, there exist m, n, p, q∈ℕ and a, b∈{±1} such that s∈(am/n, bp/q) and S∩(am/n, bp/q) = {s}. Then we may define
    f (s) = 2a+13b+15m7n11p13q.

  7. For d∈ℕ, the number of d-digit integers in S is 9d, because we have 9 choices for each digit, and all these integers are ≥10d-1. Therefore the series is bounded by

    d∈ℕ9d(10d-1)-99/100 = ∑d∈ℕ9d10-99(d-1)/100.

    This is a geometric series with ratio between successive terms 9·10-99/100; we show that this ratio is < 1. Rearranging, we find that we need to prove 1099/999 > 9, equivalently (1 + 1/9)99 > 9, which is true by the binomial series. It follows that the geometric series is convergent, and we conclude by the comparison test that the original series is convergent.